Transcript Slide 1
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یطخ لرتنک یاهمتسیس 1389 زییاپ
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What is frequency response
So far we have described the response and performance of a system in terms of complex frequency variable s=σ+jω and the location of poles and zeros in the s-plane . An important alternative approach to system analysis and design is the frequency response method .
The
frequency response
state of a system is defined as the response of the system to a steady sinusoidal input signal . We will investigate the steady-state response of the system to the sinusoidal input as the frequency varies .
When the input signal is a sinusoid , the resulting output signal for LTI systems is sinusoidal in the steady state , it differs from the input only in amplitude and phase.
Consider t he system
Y
(
s
)
T
(
s
)
R
(
s
) with If
r
(
t
)
A
sin
t
.
R
(
s
) L {
r
(
t
)}
s
2
A
2
T
(
s
) (
s
p
1 )(
s m
(
s
)
p
2 ) (
s
p n
) , where p 1 , p 2 ,…,p n are distinctive poles, then in partial fraction expansion form, we have
Y
(
s
)
s k
1
p
1
s k
2
p
2
s k n
p n
s
2
s
2 Taking the inverse Laplace transform yields
y
(
t
)
k
1
e
p
1
t
k
2
e
p
2
t
k n e
p n t
L
-
1
s
s
2 2 Suppose the system is stable , then all the poles are located in the left half plane and thus the exponential terms decay to zero as t→∞. Hence, the steady-state response of the system is
t
lim
y
(
t
) L
-
1
s
2
s
2
A T
(
j
) sin(
t
), where
T
(
j
).
Process Exposed to a Sinusoidal Input
For the system
Y
(
s
)
T
(
s
)
R
(
s
) with
r
(
t
)
A
sin
t
, the steady-state output is lim
t
y
(
t
)
A T
(
j
) sin(
t
), where
T
(
j
).
That is, the steady-state response depends only on the magnitude and phase of T(jω).
r(t) = A sin(
t)
G(s)
c(t) = |G(j
)| A sin(
t +
q
)
Advantages of the frequency response method
• • • • The sinusoidal input signal for various ranges of frequency and amplitude is readily available.
It is the most reliable and uncomplicated method for the
experimental analysis
of a system.
Control of system bandwidth as well as some measure of the response of the system to undesired noise and disturbances .
The TF describing the sinusoidal steady-state behavior of the system is easily obtained by TF.
replacing s with jω in the system
Key Components of Frequency Response Analysis
D
t p a y a c c
A r
a y a c
y s Time
D
t p
2 360 º
Graphic expression of the frequency response
1. Rectangular coordinates plot Example
G
(
s
)
2
s
10
1
G
(
j
)
10
j
2
1
G
(
j
)
0 0 .
5 1 2 3 4
G
(
j
) 10 7 .
07 4 .
47 2 .
4 1 .
64 1 .
24
G
(
j
) 0
o
45
o
63 .
435
o
75 .
964
o
80 .
538
o
82 .
875
o
5 0
.
995
84
.
29
o
10 5 1 0 0. 5 1 2 3 4 5
G
(
j
)
- 90 o
10 1
( 2
) 2
tg
1 ( 2
)
2. Polar plot The polar plot is easily useful for investigating system stability .
Example
G
(
s
)
s
(
K Ts
1 )
G
(
j
)
G
(
s
)
s
j
j
(
K j
T
1 ) The magnitude and phase response:
A
(
)
G
(
j
)
K
1
(
T
) 2 ;
(
)
G
(
j
)
[ 90
o
tg
1 (
T)] Calculate A(ω) and
( )
for different ω:
KT
Im 2
A
(
)
(
)
0
90
o
4 1 2
T KT
117
o
5 1
KT T
2
135
o
0
180
o
4KT
1
T
5
0 1 2
T -135 o
-117 o Re
The shortage of the polar plot and the rectangular coordinates plot: to synchronously investigate the cases of the lower and higher frequency band is difficult.
Idea: How to enlarge the lower frequency band and shrink (shorten) the higher frequency band
？
Bode diagram(logarithmic plots) Plot the frequency characteristic in a semilog coordinate: Magnitude response — Y-coordinate in decibels:
20 log
G
X-coordinate in logarithm of ω: logω
(
j
)
Phase response — Y-coordinate in radian:
G
(
j
)
X-coordinate in logarithm of ω: logω
Consider the general form of transfer functions.
G
(
j
)
j
k
(
j K
(
j
p
1
)((
z
1
j
)(
)
j
2
2
z
2
V
)(
n j
)
n
2 )
This may be written in Bode form by dividing through by all the constants.
G
(
j
)
j
K B
( 1
k
( 1
j
p
1
j z
1 )( 1
)( 1 2
V
j
n j z
2
)(
n
)
2 )( )
where the Bode Gain is
K B
K p
1
z
1
z n
2 2 Now consider the gain of G(j ) in dB.
G
(
j
)
dB
(
j
)
k
20 log 10
G
(
j
)
K B dB
1
j
z
1
dB
dB
1
j
p
1
dB
1
j
z
2
dB
1 2 V
j n
n
2
dB
The angle of G(j ) may be written as /
G
(
j
)
K B
1
j
z
1 1
j
z
2 (
j
)
k
1
j
p
1 1 2 V
j n
n
2 Thus it is clear that for both magnitude in dB and the angle , the total transfer function may be written in terms of the sum of its components
Frequency Response of The Typical Elements
The typical elements of the linear control systems 1. Proportional element Transfer function:
G
(
s
)
C
(
s
)
R
(
s
)
K
Frequency response:
G
(
j
)
K
G
(
j
)
K
L
(
)
20 log
G
(
j
)
20 log
K
(
)
G
(
j
)
0
o
Im
K
Polar plot
Re 0dB, 0 o
L
( ), ( )
L
(
)
20 log
K
dB
0.1 1 10 100
Bode diagram
( ) 0
o
(log )
Frequency Response of The Typical Elements
2. Integrating element Transfer function:
G
(
s
)
C
(
s
)
R
(
s
)
1
s
Frequency response:
G
(
j
)
Im 1
j
G
(
j
)
1
L
(
(
)
)
20 log
G
(
G
(
j
)
j
)
90
o
L
( ), ( )
20 log
Re
L
(
) :
20
dB
/
dec
0
0dB, 0 o 0.1 1 10 100 (log )
(
)
90
o
Polar plot Bode diagram
Frequency Response of The Typical Elements
3. Inertial element Transfer function:
G
(
j
) 1/T:
1
(
T
) 2 break frequency Im
1 1
G
(
s
)
L
(
)
20 log
(
)
C
(
s
)
R
(
s
)
tg
1
Ts
1
1 1
(
T
) 2
G
(
j
)
1
j
T
1
0
20 3
dB
log(
T
)
1
1 1 T T
T
(
T
)
0 Re
0dB, 0 o
L
( ), (
G
(
s
)
)
K T
2
s
1 :
1
20 log
K
T
0.1 1 10 100 1
T
2 (log ) 45
o
20 dB /
dec
90
o
Polar plot Bode diagram
Frequency Response of The Typical Elements
4. Oscillating element Transfer function:
G
(
j
)
G
(
j
)
L
(
)
( 1
2
T
2 )
20 log 1
maximum value
1 (
2
T G
(
j
2 ) 1
2 ( 1
2
T
2 ) 2 )
G
(
s
)
j
2
T
：
C
(
s
)
R
(
s
)
( 2
T
) 2 ( 2
T
) 2
T
2
s
2
( )
1
2
Ts
1 0
tg
1 ( 1 2
T
2
T
0
20 40 log( log(
2
T
) )
n
(
1
T
n n
)
2 )
1 Make:
d d
(
G
(
j
) )
0
M r r
resonant p eak
M r
r
n G
(
j
r
) 1
2
2
2 (0
1 1
2
2 2 )
The polar plot and the Bode diagram:
Im
0 1 Re
L
( ), ( )
n
r
1 /
T
20 log
M r
20 log( 1 2
)
0dB, 0 o 10 100 (log ) 0.1 1
j
1 2
(
n
)
90
o
40 dB /
dec
Polar plot
180
o
1 .
2 .
0
r
2
2
(
r
n
n
)
r
M r
Bode diagram
0
unstable s ystem No resonan ce
, Optimal Second order System
•
Second-Order System
•
The transfer function of a 2 nd -order system:
H
(
s
)
s
2 2
n
2
n s
n
2 •
The frequency response of this system can be modeled as:
H
(
j
)
dB
20 log
n
1 2
n
2 2 2
n
2
H
(
j
)
dB
20 log 2
n
2 2
H
(
j
) 40 log
n
tan 1 2
n
180
40 dB/decade Changes by
Frequency Response of The Typical Elements
5. Differentiating element Transfer function:
G
(
s
)
Ts
2
Ts
s
1 2
Ts
1
differenti al first
order differenti al second
order differenti al
Im Im Im
differential
Re 1 Re 1 Re
1th-order differential Polar plot 2th-order differential
Frequency Response of The Typical Elements
Because of the transfer functions of the differentiating elements are the reciprocal of the transfer functions of Integrating element, Inertial element and Oscillating element respectively, that is:
s Ts
1
T
2
s
2
2
Ts
1
inverse
inverse
inverse
1
s
1
Ts
1 1
T
2
s
2
2
Ts
1 the Bode curves of the differentiating elements are symmetrical to the logω-axis with the Bode curves of the Integrating element, Inertial element and Oscillating element respectively.
Then we have the Bode diagram of the differentiating elements:
Frequency Response of The Typical Elements
L
( ), ( )
L
( ), ( ) 180
o
0dB, 0 o
L
( ) : 20
dB
/
dec
0.1 1 10
differential
100 ( ) 90
o
(log ) 90
o
0dB, 0 o 0.1
n
1
T
1 10
0dB, 0 o
90
o
45
o
0.1
L
( ), ( ) 20
dB
/
dec
(log ) 1 10 100
1th-order differential
100 40
dB
/
dec
(log ) 20 log( 1 2 ) 20 log
M r
2th-order differential
Frequency Response of The Typical Elements
6. Delay element Transfer function:
G
(
s
)
C
(
s
)
R
(
s
)
e
s G
(
j
)
e
j
Im
G
( (
j
)
)
1
G
(
j
L
(
) )
0
L
( ), ( )
R=1
0dB, 0 o Re 0.1 1 10 100 (log )
Polar plot Bode diagram
method to plot the magnitude response of the Bode diagram
•
Transfer function:
H
(
s
) (
s
1000 (
s
10 )(
s
2 ) 50 )
H
(
j
) (
j
1000 (
j
10 )(
j
2 ) 50 )
j
4 10
j
2 1
j
1 50 1 •
The critical frequencies are
= 2 (zero), 10 (pole), and 50 (pole). •
MATLAB (exact resp.):
w = logspace(-1,3,300); s = j*w; H = 1000*(s+2)./(s+10)./(s+50); magdB = 20*log10(abs(H)); phase = angle(H)*180/pi; •
MATLAB (Bode):
num = [1000 2000]; den = conv([1 1o], [1 50]); bode(num, den); •
Bode plots are useful as an analytic tool.
Example: Comparison of Exact and Bode Plots
method to plot the magnitude response of the Bode diagram
G
(
s
)
s
( 10 (
s
s
5 )( 10 )
s
2 ) Bode Form:
G
(
j
)
j
( 10 (
j
j
5 )( 10 )
j
2 )
j
( 1 10
j
( 1
j
/ 5 )( 1 / 10 )
j
/ 2 ) Plot the asymptotic approximations for each term separately, for both magnitude and angle.
Then add them together to get the system asymptotic approximation.
Sketch in the Bode plot curve.
j
( 1 10
j
( 1 / 5
j
)( 1 / 10
j
) / 2 )
60 40 20 0 -20 -40 -60 1
j
-90 -100 -120 -140 -160 -180 10 -2 10 -1 .2
.5
10 0 2
Frequency (rad/sec)
5 10 1 1
j
/ 10 1
j
1
/ 5 1
j
1
/ 2 10 2
method to plot the magnitude response of the Bode diagram
method to plot the magnitude response of the Bode diagram
G
(
s
)
s
5 1
s
2 100 2
s
100 Bode Form:
G
(
j
)
j
5 1 ( ) 2 100 2
j
100 1
j
5 / 1 1 0 .
2
j
1 / 10 ( / 10 ) 2 The damping ratio for the second-order term is = 0.1 and the natural frequency is 10 rad./s .
G
(
j
) 1
j
5
j
5 1 / 1 ( ) 1 2 100 2
j
0 .
2
j
1 / 10 100 ( / 10 ) 2
14 -20dB/dec.
0 14dB -20 -40 -60 0 -50 -100 -150 -200 -270 -250 10 -1 -45
/dec.
-135
/dec.
10 0 p = 1
n
10 1 = 10
Frequency (rad/sec)
-60dB/dec.
r
n
1 2 2
M r
2 10 1 1 1 2 2 0 .
1 2 5 .
05 9 .
9 14 .
6
dB
-90
/dec.
10 2
method to plot the magnitude response of the Bode diagram
G
(
s
)
10 (
s
1 )
s
( 0 .
1
s
1 )( 0 .
01 2
s
2
0 .
01
s
1 )
20dB, 45 o 0dB, 0 o 0.1 -20dB, -45 o
L
( 1 ), ( ) 40dB, 90 o －
20dB/dec
10 -40dB, -90 o -60dB.-135 o -80dB,-180 o -100dB,-225 o -120dB,-270 o －
20dB/dec
100
r
1.25dB
(log ) －
60dB/dec There is a resonant peak M
r
at:
M r
2
r
n
1 1
2
1 .
154
1 .
25
dB
100 1
2
2 1
2
0 .
5 2
70 .
7